Question

if 2^a=3^b=6^c then show that c=ab/a+b

Answer 1

Note:

★ a^m × a^n = a^(m + n)

★ a^m / a^n = a^(m - n)

★ a^m × b^m = (a×b)^m

★ a^m / b^m = (a/b)^m

★ [ a^m ]^n = a^(m×n)

★ a^m = b => a = b^(1/m)

Solution:

• Given : 2^a = 3^b = 6^c
• To prove : c = ab/(a + b)

Let 2^a = 3^b = 6^c = k .

Thus,

• If 2^a = k , then

2 = k^(1/a)

• If 3^b = k , then

3 = k^(1/b)

• If 6^c = k , then

6 = k^(1/c)

Now ,

=> 6 = k^(1/c)

=> 2×3 = k^(1/c)

=> [k^(1/a)] × [k^(1/b)] = k^(1/c)

=> k^(1/a + 1/b) = k^(1/c)

=> 1/a + 1/b = 1/c

=> 1/c = 1/a + 1/b

=> 1/c = (b + a)/ab

=> c = ab/(a + b)

Hence proved .

Answer 2

Answer:

Given : 2^a = 3^b = 6^c

To prove : c = ab/(a + b)

Let 2^a = 3^b = 6^c = k .

Thus,

• If 2^a = k , then

2 = k^(1/a)

• If 3^b = k , then

3 = k^(1/b)

• If 6^c = k , then

6 = k^(1/c)

Now ,

=> 6 = k^(1/c)

=> 2×3 = k^(1/c)

=> [k^(1/a)] × [k^(1/b)] = k^(1/c)

=> k^(1/a + 1/b) = k^(1/c)

=> 1/a + 1/b = 1/c

=> 1/c = 1/a + 1/b

=> 1/c = (b + a)/ab

=> c = ab/(a + b)

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Step-by-step explanation: